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Check if row's date range overlap any previous rows date range in Python / Pandas Dataframe

I have some data in a pandas dataframe that contains a rank column, a start date and an end date. The data is sorted on the rank column lowest to highest (consequently the start/end dates are unordered). I wish to remove every row whose date range overlaps ANY PREVIOUS rows'
By way of a toy example:
Raw Data
Rank Start_Date End_Date
1 1/1/2021 2/1/2021
2 1/15/2021 2/15/2021
3 12/7/2020 1/7/2021
4 5/1/2020 6/1/2020
5 7/10/2020 8/10/2020
6 4/20/2020 5/20/2020
Desired Result
Rank Start_Date End_Date
1 1/1/2021 2/1/2021
4 5/1/2020 6/1/2020
5 7/10/2020 8/10/2020
Explanation: Row 2 is removed because its start overlaps Row 1, Row 3 is removed because its end overlaps Row 1. Row 4 is retained as it doesn’t overlap any previously retained Rows (ie Row 1). Similarly, Row 5 is retained as it doesn’t overlap Row 1 or Row 4. Row 6 is removed because it overlaps with Row 4.
Attempts:
I can use np.where to check the previous row with the current row and create a column “overlap” and then subsequently filter this column. But this doesn’t satisfy my requirement (ie in the toy example above Row 3 would be included as it doesn’t overlap with Row2 but should be excluded as it overlaps with Row 1).
df['overlap'] = np.where((df['start']> df['start'].shift(1)) &
(df['start'] < df['end'].shift(1)),1 ,0)
df['overlap'] = np.where((df['end'] < df['end'].shift(1)) &
(df['end'] > df['start'].shift(1)), 1, df['overlap'])
I have tried an implementation based on answers from this question Removing 'overlapping' dates from pandas dataframe, using a lookback period from the End Date, but the length of days between my Start Date and End Date are not constant, and it doesnt seem to produce the correct answer anyway.
target = df.iloc[0]
day_diff = abs(target['End_Date'] - df['End_Date'])
day_diff = day_diff.reset_index().sort_values(['End_Date', 'index'])
day_diff.columns = ['old_index', 'End_Date']
non_overlap = day_diff.groupby(day_diff['End_Date'].dt.days // window).first().old_index.values
results = df.iloc[non_overlap]

Two intervals overlap if (a) End2>Start1 and (b) Start2<End1:
We can use numpy.triu to calculate those comparisons with the previous rows only:
a = np.triu(df['End_Date'].values>df['Start_Date'].values[:,None])
b = np.triu(df['Start_Date'].values<df['End_Date'].values[:,None])
The good rows are those that have only True on the diagonal for a&b
df[(a&b).sum(0)==1]
output:
Rank Start_Date End_Date
1 2021-01-01 2021-02-01
4 2020-05-01 2020-06-01
5 2020-07-10 2020-08-10
NB. as it needs to calculate the combination of rows, this method can use a lot of memory when the array becomes large, but it should be fast

Another option, that could help with memory usage, is a combination of IntervalIndex and a for loop:
Convert dates:
df.Start_Date = df.Start_Date.transform(pd.to_datetime, format='%m/%d/%Y')
df.End_Date = df.End_Date.transform(pd.to_datetime, format='%m/%d/%Y')
Create IntervalIndex:
intervals = pd.IntervalIndex.from_arrays(df.Start_Date,
df.End_Date,
closed='both')
Run a for loop (this avoids broadcasting, which while fast, can be memory intensive, depending on the array size):
index = np.arange(intervals.size)
keep = [] # indices of `df` to be retained
# get rid of the indices where the intervals overlap
for interval in intervals:
keep.append(index[0])
checks = intervals[index].overlaps(intervals[index[0]])
if checks.any():
index = index[~checks]
else:
break
if index.size == 0:
break
df.loc[keep]
Rank Start_Date End_Date
0 1 2021-01-01 2021-02-01
3 4 2020-05-01 2020-06-01
4 5 2020-07-10 2020-08-10

find max-min values for one column based on another

I have a dataset that looks like this.
datetime id
2020-01-22 11:57:09.286 UTC 5
2020-01-22 11:57:02.303 UTC 6
2020-01-22 11:59:02.303 UTC 5
Ids are not unique and give different datetime values. Let's say:
duration = max(datetime)-min(datetime).
I want to count the ids for what the duration max(datetime)-min(datetime) is less than 2 seconds. So, for example I will output:
count = 1
because of id 5. Then, I want to create a new dataset which contains only those rows with the min(datetime) value for each of the unique ids. So, the new dataset will contain the first row but not the third. The final data set should not have any duplicate ids.
datetime id
2020-01-22 11:57:09.286 UTC 5
2020-01-22 11:57:02.303 UTC 6
How can I do any of these?
P.S: The dataset I provided might not be a good example since the condition is 2 seconds but here it's in minutes

Do you want this? :
df.datetime = pd.to_datetime(df.datetime)
c = 0
def count(x):
global c
x = x.sort_values('datetime')
if len(x) > 1:
diff = (x.iloc[-1]['datetime'] - x.iloc[0]['datetime'])
if diff < timedelta(seconds=2):
c += 1
return x.head(1)
new_df = df.groupby('id').apply(count).reset_index(drop=True)
Now, if you print c it'll show the count which is 1 for this case and new_df will hold the final dataframe.

How to find missing date rows in a sequence using pandas?

I have a dataframe with more than 4 million rows and 30 columns. I am just providing a sample of my patient dataframe
df = pd.DataFrame({
'subject_ID':[1,1,1,1,1,2,2,2,2,2,3,3,3],
'date_visit':['1/1/2020 12:35:21','1/1/2020 14:35:32','1/1/2020 16:21:20','01/02/2020 15:12:37','01/03/2020 16:32:12',
'1/1/2020 12:35:21','1/3/2020 14:35:32','1/8/2020 16:21:20','01/09/2020 15:12:37','01/10/2020 16:32:12',
'11/01/2022 13:02:31','13/01/2023 17:12:31','16/01/2023 19:22:31'],
'item_name':['PEEP','Fio2','PEEP','Fio2','PEEP','PEEP','PEEP','PEEP','PEEP','PEEP','Fio2','Fio2','Fio2']})
I would like to do two things
1) Find the subjects and their records which are missing in the sequence
2) Get the count of item_name for each subjects
For q2, this is what I tried
df.groupby(['subject_ID','item_name']).count() # though this produces output, column name is not okay. I mean why do it show the count value on `date_visit` column?
For q1, this is what I am trying
df['day'].le(df['shift_date'].add(1))
I expect my output to be like as shown below

You can get the first part with:
In [14]: df.groupby("subject_ID")['item_name'].value_counts().unstack(fill_value=0)
Out[14]:
item_name Fio2 PEEP
subject_ID
1 2 3
2 0 5
3 3 0
EDIT:
I think you've still got your date formats a bit messed up in your sample output, and strongly recommend switching everything to the ISO 8601 standard since that prevents problems like that down the road. pandas won't correctly parse that 11/01/2022 entry on its own, so I've manually fixed it in the sample.
Using what I assume these dates are supposed to be, you can find the gaps by grouping and using .resample():
In [73]: df['dates'] = pd.to_datetime(df['date_visit'])
In [74]: df.loc[10, 'dates'] = pd.to_datetime("2022-01-11 13:02:31")
In [75]: dates = df.groupby("subject_ID").apply(lambda x: x.set_index('dates').resample('D').first())
In [76]: dates.index[dates.isnull().any(axis=1)].to_frame().reset_index(drop=True)
Out[76]:
subject_ID dates
0 2 2020-01-02
1 2 2020-01-04
2 2 2020-01-05
3 2 2020-01-06
4 2 2020-01-07
5 3 2022-01-12
6 3 2022-01-14
7 3 2022-01-15
You can then add seq status to that first frame by checking whether the ID shows up in this new frame.

pandas - How to check consecutive order of dates and copy groups of them?

At first I have two problems, the first will follow now:
I a dataframe df with many times the same userid and along with it a date and some unimportant other columns:
userid date
0 243 2014-04-01
1 234 2014-12-01
2 234 2015-11-01
3 589 2016-07-01
4 589 2016-03-01
I am currently trying to groupby them by userid and sort the dates descending and cut out the twelve oldest. My code looks like this:
df = df.groupby(['userid'], group_keys=False).agg(lambda x: x.sort_values(['date'], ascending=False, inplace=False).head(12))
And I get this error:
ValueError: cannot copy sequence with size 6 to array axis with dimension 12
At the moment my aim is to avoid to split the dataframe in individual ones.
My second problem is more complex:
I try to find out if the sorted dates (respectively per group of userids) are monthly consecutive. This means if there is an date for one group of userid, for example userid: 234 and date: 2014-04-01, the next entry below must be userid: 234 and date:2014-03-01. There is no focus on the day, only the year and month are important.
And only this consecutive 12 dates should be copied in another dataframe.
A second dataframe df2 contains the same userid, but they are unique and another column is 'code'. Here is an example:
userid code
0 433805 1
24 5448 0
48 3434 1
72 34434 1
96 3202 1
120 23766 1
153 39457 0
168 4113 1
172 3435 5
374 34093 1
I summarize: I try to check if there are 12 consecutive months per userid and copy every correct sequence in another dataframe. For this I have also compare the 'code' from df2.
This is a version of my code:
df['YearMonthDiff'] = df['date'].map(lambda x: 1000*x.year + x.month).diff()
df['id_before'] = df['userid'].shift()
final_df = pd.DataFrame()
for group in df.groupby(['userid'], group_keys=False):
fi = group[1]
if (fi['userid'] <> fi['id_before']) & group['YearMonthDiff'].all(-1.0) & df.loc[fi.userid]['code'] != 5:
final_df.append(group['userid','date', 'consum'])
At first calculated from the date an integer and made diff(). On other posts I saw they shift the column to compare the values from the current row and the row before. Then I made groupby(userid) to iterate over the single groups. Now it's extra ugly I tried to find the beginning of such an userid-group, try to check if there are only consecutive months and the correct 'code'. And at least I append it on the final dataframe.
On of the biggest problems is to compare the row with the following row. I can iterate over them with iterrow(), but I cannot compare them without shift(). There exits a calendar function, but on these I will take a look on the weekend. Sorry for the mess I am new to pandas.
Has anyone an idea how to solve my problem?

for your first problem, try this
df.groupby(by='userid').apply(lambda x: x.sort_values(by='date',ascending=False).iloc[[e for e in range(12) if e <len(x)]])

Using groupby and nlargest, we get the index values of those largest dates. Then we use .loc to get just those rows
df.loc[df.groupby('userid').date.nlargest(12).index.get_level_values(1)]
Consider the dataframe df
dates = pd.date_range('2015-08-08', periods=10)
df = pd.DataFrame(dict(
userid=np.arange(2).repeat(4),
date=np.random.choice(dates, 8, False)
))
print(df)
date userid
0 2015-08-12 0 # <-- keep
1 2015-08-09 0
2 2015-08-11 0
3 2015-08-15 0 # <-- keep
4 2015-08-13 1
5 2015-08-10 1
6 2015-08-17 1 # <-- keep
7 2015-08-16 1 # <-- keep
We'll keep the latest 2 dates per user id
df.loc[df.groupby('userid').date.nlargest(2).index.get_level_values(1)]
date userid
0 2015-08-12 0
3 2015-08-15 0
6 2015-08-17 1
7 2015-08-16 1

Maybe someone is interested, I solved my second problem thus:
I cast the date to an int, calculated the difference and I shift the userid one row, like in my example. And then follows this... found a solution on stackoverflow
gr_ob = df.groupby('userid')
gr_dict = gr_ob.groups
final_df = pd.DataFrame(columns=['userid', 'date', 'consum'])
for group_name in gr_dict.keys():
new_df = gr_ob.get_group(group_name)
if (new_df['userid'].iloc[0] <> new_df['id_before'].iloc[0]) & (new_df['YearMonthDiff'].iloc[1:] == -1.0).all() & (len(new_df) == 12):
final_df = final_df.append(new_df[['userid', 'date', 'consum']])

in Pandas Dataframe, after I set an entire column, I can't update another column with times

I have a large dataframe. One column is a timestamp, and another is boolean. When I set the entire boolean column at once, I can no longer update anything in the timestamp column - when I try, it doesn't complain, but the value doesn't change. Here's a simplified example:
start = pd.to_datetime('20140401')
df = pd.DataFrame(index=pd.DateRange(start,periods=1), columns=['timenow','Live'])
df.at[start,'timenow'] = datetime.today() # initial value
print(df)
df.at[start,'timenow'] = datetime.today() # this works
print(df)
df.Live = True
df.at[start,'timenow'] = datetime.today() # this doesn't work - nothing changes
print(df)
I would expect to see 3 distinct times, but instead the 2nd time stays put when I try to change it:
timenow Live
2014-04-01 2014-04-24 01:33:30.037108 NaN
[1 rows x 2 columns]
timenow Live
2014-04-01 2014-04-24 01:33:30.040039 NaN
[1 rows x 2 columns]
timenow Live
2014-04-01 2014-04-24 01:33:30.040039 True
[1 rows x 2 columns]
What am I missing?

You are probably working on the a view rather than the data directly, this should work:
df.loc[start,'timenow']= datetime.today()
print df